Hard Projectile Motion Question

[emmettecox]

Homework Statement

A cannon is firing from an elevated position 100 m above a target 800 m away. The ititial velocity of the cannon is 500 m/s.

Homework Equations

Vx = VCOSq
Vy = VSINq

a = (Vf - Vi)/(t)
Vf = Vi + a t

s = 0.5 (Vo +Vf)t

Vf2 = Vo2 + 2aS

S = Vit + 0.5at2

The Attempt at a Solution

What I tried was dividing the problem in two pieces and working backwards. The angle of reach to get to a position level with the cannon is theta = 0.5(arcsin (gx/v^2)) So if I can get the distance from the target where the cannonball is level with cannon I can go back and get the original angle. At that level point the velocity should be the same as when it left the cannon but going down.

The time to get to the target should be the distance in the x direction from that point Xb times Vx = t = Xb/(VxCOS q). The time of flight should be found out by calculating s = -VSINq t + (0.5)at^2 and the other kinematics equations. But I keep going around in cirles on this.

 

[rootX]

Homework Statement

A cannon is firing from an elevated position 100 m above a target 800 m away. The ititial velocity of the cannon is 500 m/s.

Homework Equations

Vx = VCOSq
Vy = VSINq

a = (Vf - Vi)/(t)
Vf = Vi + a t

s = 0.5 (Vo +Vf)t

Vf2 = Vo2 + 2aS

S = Vit + 0.5at2

The Attempt at a Solution

What I tried was dividing the problem in two pieces and working backwards. The angle of reach to get to a position level with the cannon is theta = 0.5(arcsin (gx/v^2)) So if I can get the distance from the target where the cannonball is level with cannon I can go back and get the original angle. At that level point the velocity should be the same as when it left the cannon but going down.

The time to get to the target should be the distance in the x direction from that point Xb times Vx = t = Xb/(VxCOS q). The time of flight should be found out by calculating s = -VSINq t + (0.5)at^2 and the other kinematics equations. But I keep going around in cirles on this.

What are you trying to find?

Here's another method:

find y-x relationship
with intial pos of (0,0) and final of (800,-100)
this is a direct (wouldn't say easier) way to find theta

 

[tiny-tim]

So if I can get the distance from the target where the cannonball is level with cannon I can go back and get the original angle.

Hi emmettecox!

I don't understand why you think this will help … it won't!

Homework Equations

Vx = VCOSq
Vy = VSINq

a = (Vf - Vi)/(t)
Vf = Vi + a t

s = 0.5 (Vo +Vf)t

Vf2 = Vo2 + 2aS

S = Vit + 0.5at2

ok, but it's no good just writing the general equations without
being able to apply them.

Hint: Which of those equations shoud have Vx in them, and which Vy?

 

[emmettecox]

What I am trying to find is the equation for the angle of reach for a projectile fired from an elevated position. If you fire a projectile at an angle q from the same height as your target and you know the distance, x to your target and the initial velocity v you get

q = 0.5[arcsin(gx/v^2)

If you are firing from a height h above the target your time of flight is longer. I worked on an equation and got q = 0.5[arccos{sqrt(1+gx^2)/(v^2)}] But I've been all over the place trying to solve this simplifying it using trig identities. You can solve it piece by piece eaisly enough, but I was trying for one equation for the angle.

I could use some help if anyone has done this problem this way.

 

[GoldPheonix]

Homework Statement

A cannon is firing from an elevated position 100 m above a target 800 m away. The ititial velocity of the cannon is 500 m/s.

Homework Equations

Vx = VCOSq
Vy = VSINq

a = (Vf - Vi)/(t)
Vf = Vi + a t

s = 0.5 (Vo +Vf)t

Vf2 = Vo2 + 2aS

S = Vit + 0.5at2

The Attempt at a Solution

What I tried was dividing the problem in two pieces and working backwards. The angle of reach to get to a position level with the cannon is theta = 0.5(arcsin (gx/v^2)) So if I can get the distance from the target where the cannonball is level with cannon I can go back and get the original angle. At that level point the velocity should be the same as when it left the cannon but going down.

The time to get to the target should be the distance in the x direction from that point Xb times Vx = t = Xb/(VxCOS q). The time of flight should be found out by calculating s = -VSINq t + (0.5)at^2 and the other kinematics equations. But I keep going around in cirles on this.

Honestly, this is not a hard problem at all. Well, it will seem that way, until you get used to approaching story problems in a ritualized way.


So, let's start off with what you know. You have some object, which is being shot out of a cannon. All right, so let's talk about systematically approaching problems.

In any given problem, you're dealing with initial conditions, equations, and then your predictions. Equations are used to predict phenomena, from things as large as galaxies consuming one another in their gravitational pull to quantum particles interacting to create the color of light that shines off of a piece of metal. Fortunately, you're not doing anything that complex; in fact, all you have to do right now is some algebra. So, in short, the systematic way to approach a problem is this:

Take known initial conditions (initial velocity, acceleration due to gravity, initial height, et cetera), plug them into an equation, find out what predictions you can make given these initial conditions (in mathematical terms, use algebra to find out unknowns).


So, what are our knowns/initial conditions?

S = 500 m/s at O (theta) degrees

So, you're trying to predict what angle you'd need to get that cannon ball to reach that point across the field. At least, I assume as much, because it's the only real unknown here.


All right, this is a free fall problem, so we'll use the known free fall equations:

x(t) = u*t + x_initial
y(t) = -.5gt^2 + v*t +y_initial
v = S*singO
u = S*cosO


NOTE: I'm going to call the speed in the x-direction "u" and the speed in the y-direction "v". Thus they get the appropriate trig function attached to them. "S" is, from above, the speed of 500 m/s.


So, we take these equations and we simplify them. Your initial height and distance outward are zero. Now, you've got fairly simple set of equations to work with.

So, when I write the term "x(t)", what I mean is some mathematical expression evaluated for some time, "t", and this expression happens to equal the number corresponding to the particular x-position where the cannon ball is.

So, if I think of some final "time" when it reaches some final position, even if at the moment I don't know what the time is, I can postulate that there is in fact a corresponding final x-position which is equal to the mathematical expression evaluated at that "final" time. IE: "x_final = x(t_final)"

We don't know what final time is right now, but we do know what that final position is. It's 800 meters to our right. (I'm no allowed to solve this problem for you, but I will give you some symbolic variables, and allow you to play on your own)

x_final = t_final*u

y_final = -1/2g(t_final)^2 + v(t_final)


Now, both the "u" and the "v" have the unknown angle placed directly into them:

x_final = S*cosO*(t_final)

y_final = -1/2g(t_final)^2 + S*sinO*(t_final)


You need to know "O". Well, there's a popular trick in both math and physics. If you don't know two things, well, if you have two equations, and you solve them both down for the one particular value (In this case, t_final), well, you know then that if they both equal t_final, then the two mathematical statements equal one another. Then you're just left with the other unknown, which you can solve for.


So, let's just give you an example of this:

t_final = S*cosO
t_final = whatever you get from the quadratic equation

Then see if you can follow my process.

 

[tiny-tim]

What I am trying to find is the equation for the angle of reach for a projectile fired from an elevated position.
…
You can solve it piece by piece eaisly enough, but I was trying for one equation for the angle.

Hi emmettecox! smile:

I haven't tried to work it out myself, but your equation q = 0.5[arccos{sqrt(1+gx^2)/(v^2)}] can't be correct, since the dimensions are wrong.

gx² is L³/T², but v² is L²/T² (and 1 is just a number).